Pandas 和 Sets - ValueError：值的长度与索引的长度不匹配
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Pandas and Sets - ValueError: Length of values does not match length of index
提问by johnaco
I am trying to create a new column in my dataframe that contains the intersection of two sets (each contained in two separate columns). The columns themselves hold sets.
我正在尝试在我的数据框中创建一个新列，该列包含两个集合的交集（每个都包含在两个单独的列中）。列本身包含集合。
dfc['INTERSECTION'] = set(dfc.TABS1).intersection(set(dfc.TABS2))
I get a Value error. I was able to do
我收到一个值错误。我能够做到
dfc['LEFT'] = set(dfc.TABS1) - set(dfc.TABS2)
no problem. TABS1 and TABS2 have values.
没问题。TABS1 和 TABS2 有值。
Any thoughts? Thanks.
有什么想法吗？谢谢。
I am adding example data below.
我在下面添加示例数据。
GROUP TABS1 TABS2
A {'T1','T2','T3'} {'T2','T3','T4'}
B {'T5', 'T6'} {'T6'}
Chris gave example, but using very different data set. I am looking for the intersection of TAB1 and TAB2 in a third column 'INTERSECTION. As mentioned above, I have no problems with
Chris 举了例子，但使用了非常不同的数据集。我正在第三列“INTERSECTION”中寻找 TAB1 和 TAB2 的交集。如上所述，我没有问题
dfc['LEFT'] = set(dfc.TAB1) - set(dfc.TAB2)
This looks like it should be so straight forward...
这看起来应该如此简单......
回答by Yo_Chris
setremoves duplicates so you end up with a dict with a length less than the length of your dataframe. You need make sure the length of the array you are assign to a new column is equal to the length of the dataframe. You can replace the non-intersections with NaNif you want using list comprehension:
set删除重复项，因此您最终会得到一个长度小于数据帧长度的字典。您需要确保分配给新列的数组长度等于数据帧的长度。NaN如果您想使用列表理解，您可以将非交集替换为：
# sample data
df = pd.DataFrame([[1,2,3], [1,2,3], [2,3,4], [3,4,5]], columns=list('abc'))
# list comprehension
df['intersection'] = [a if a in set(df['b']) else np.nan for a in df['a']]
a b c intersection
0 1 2 3 NaN
1 1 2 3 NaN
2 2 3 4 2.0
3 3 4 5 3.0
